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Physics 207: Lecture 7, Pg 1 Physics 207, Lecture 7, Sept. 25 Agenda: Assignment: l WebAssign Problem Set 3 due Tuesday midnight l MidTerm Thursday, Oct.

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Presentation on theme: "Physics 207: Lecture 7, Pg 1 Physics 207, Lecture 7, Sept. 25 Agenda: Assignment: l WebAssign Problem Set 3 due Tuesday midnight l MidTerm Thursday, Oct."— Presentation transcript:

1 Physics 207: Lecture 7, Pg 1 Physics 207, Lecture 7, Sept. 25 Agenda: Assignment: l WebAssign Problem Set 3 due Tuesday midnight l MidTerm Thursday, Oct. 5, Chapters 1-6, 90 minutes, 7-8:45 PM l NOTE: Assigned Rooms are 105 and 113 Psychology l Chapter 5 (Forces and Newton’s Laws)   Static Friction  Problem exercise Chapter 6 (Circular Motion and Other Applications)  Friction (a external force that opposes motion)  Uniform and non-uniform circular motion  Accelerated Frames  Resistive Forces

2 Physics 207: Lecture 7, Pg 2 Friction l What does it do?  It opposes motion! l How do we characterize this in terms we have learned?  Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then implied)! amaama F F APPLIED f f FRICTION gmggmg N i j See text: 5.8

3 Physics 207: Lecture 7, Pg 3 Model for Sliding Friction (with motion) N l The direction of the frictional force vector is perpendicular to the normal force vector N. f N l The magnitude of the frictional force vector |f K | is proportional to the magnitude of the normal force |N |. fN g  |f K | =  K | N | ( =  K  |  mg | in the previous example)  The “heavier” something is, the greater the frictional force The constant  K is called the “coefficient of kinetic friction”. l Depending on the other forces speed may increase or decrease See text: 6-1

4 Physics 207: Lecture 7, Pg 4 Case study... l Dynamics: x-axis i :ma x = F  K N y-axis j : ma y = 0 = N – mg or N = mg soF  K mg = m a x axmaxaxmax F gmggmg N i j  K mg See text: 6-1 v fkfk fkfk

5 Physics 207: Lecture 7, Pg 5 Lecture 7, Example 1 Friction and Motion A box of mass m 1 = 1 kg is being pulled by a horizontal string having tension T = 30 N. It slides with friction (  k = 0.5) on top of a second box having mass m 2 = 2 kg, which in turn slides on an ice rink (frictionless). Let g = 10 m/s 2  What is the acceleration of the second box ? 1.Focus first on the top block 2. Find frictional force and use action/reaction force pairs 3. Then discuss the second block (A) a = 0 m/s 2 (B) a = 2.5 m/s 2 (C) a = 10 m/s 2 m2m2m2m2 T m1m1m1m1  slides with friction (  k =0.5  ) slides without friction a = ? v

6 Physics 207: Lecture 7, Pg 6 l Finally, solve F x = ma in the horizontal direction: m2m2 f f 2,1 = -  K m 1 g  K m 1 g = m 2 a Lecture 7, Exercise 1 Solution 2.5 m/s 2 to the left

7 Physics 207: Lecture 7, Pg 7 Lecture 7, Exercise 2 Incline dynamics A block of mass m, is placed on a rough inclined plane (  > 0) and given a brief push. It motion thereafter is down the plane with a constant speed.  If a similar block (same  ) of mass 2m were placed on the same incline and given a brief push with v 0 down the block, it will (A) (A) decrease its speed (B) increase its speed (C) (C) move with constant speed m

8 Physics 207: Lecture 7, Pg 8 Lecture 7, Exercise 2 Solution l Draw FBD and find the total force in the x-direction  i j 2 mg N  KNKN F TOT,x = 2mg sin  K 2mg cos  = 2 ma = ma = 0 (case when just m) Doubling the mass will simply double both terms…net force will still be zero ! Speed will still be constant ! (C)

9 Physics 207: Lecture 7, Pg 9 Static Friction... F F net gmggmg N i j fSfS See text: Ch 5.8 l So far we have considered friction acting when something has a non-zero velocity  We also know that it acts in fixed or “static” systems  In general there is a second parameter, the coefficient of static friction or  S. In these cases, the force provided by friction depends on the forces applied to the system (with f s ≤  S N) Opposes motion (i.e., acceleration) that would occur if  S were zero

10 Physics 207: Lecture 7, Pg 10 Static Friction... l Opposes motion except here a = 0 is the constaint i :F net  f S = 0 j :N = mg i j F net gmggmg N fSfS While the block is static: f S  F net (unlike kinetic friction) l f s is NOT fixed in magnitude See text: Ch 5.8

11 Physics 207: Lecture 7, Pg 11 Static Friction... F gmggmg N i j fSfS The maximum possible force that the friction between two objects can provide is f MAX =  S N, where  s is the “coefficient of static friction”.  So f S   S N.  As one increases F, f S gets bigger until f S =  S N and the object “breaks loose” and starts to move. See text: Ch 5.8

12 Physics 207: Lecture 7, Pg 12 Static Friction... F  S is discovered by increasing F until the block starts to slide: i :F MAX  S N = 0 j :N = mg  S  F MAX / mg F F MAX gmggmg N i j  S mg See text: Ch 5.8 Active Figure

13 Physics 207: Lecture 7, Pg 13 Additional comments on Friction: See text: 6-1 Since f =  N, the force of friction does not depend on the area of the surfaces in contact (this is not strictly true, for example narrow tires reduce rolling resistance). Logic dictates that  S >  K for any system

14 Physics 207: Lecture 7, Pg 14 Coefficients of Friction Material on Material  s = static friction  k = kinetic friction steel / steel add grease to steel metal / ice brake lining / iron tire / dry pavement tire / wet pavement0.80.7

15 Physics 207: Lecture 7, Pg 15 Lecture 7, Exercise 3 Friction and Motion A box of mass m 1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 30 N. This box (1) is on top of a second box of mass m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are  s =3.5 and  k = 0.5. The second box can slide freely (frictionless) on an ice rink surface. The acceleration of box 1 is (A) Greater than (B) Equal to (C) Smaller than the acceleration of box 2 ? m2m2m2m2 T m1m1m1m1  friction coefficients  s =3.5 and  k =0.5 slides without friction a2a2 a1a1

16 Physics 207: Lecture 7, Pg 16 Newton’s Laws and Circular Motion (Chapter 6) v R Centripedal Acceleration a C = v 2 /R What is Centripedal Force ? F C = ma C = mv 2 /R aCaC Animation

17 Physics 207: Lecture 7, Pg 17 Applications Mass based separations: Centrifuges Mass Spectroscopy How many g’s? a c =v 2 / r and f = 10 4 rpm is typical with r = 0.1 m and v =  r = 2  f r v = (2  x 10 4 / 60) x 0.1 m/s =100 m/s a c = 1 x 10 4 / 0.1 m/s 2 = g’s Before After

18 Physics 207: Lecture 7, Pg 18 Lecture 7, Example 4 Circular Motion Forces with Friction (recall ma C = m v 2 / R F f ≤  s N ) l How fast can the race car go ? (How fast can it round a corner with this radius of curvature?) m car = 1600 kg  S = 0.5 for tire/road R = 80 m g = 10 m/s 2 R (A) 10 m/s (B) 20 m/s (C) 75 m/s (D) 750 m/s

19 Physics 207: Lecture 7, Pg 19 Banked Corners In the previous scenario, we drew the following free body diagram for a race car going around a curve on a flat track. N mgmg FfFf What differs on a banked curve ?

20 Physics 207: Lecture 7, Pg 20 Banked Corners Free Body Diagram for a banked curve. Use rotated x-y coordinates Resolve into components parallel and perpendicular to bank N mgmg FfFf For very small banking angles, one can approximate that F f is parallel to ma. This is equivalent to the small angle approximation sin  = tan . macmaci j 

21 Physics 207: Lecture 7, Pg 21 Non uniform Circular Motion Earlier we saw that for an object moving in a circle with non uniform speed then a = a r + a t (radial and tangential) arar atat What are F r and F t ? ma r and ma t

22 Physics 207: Lecture 7, Pg 22 Lecture 7, Example 5 Gravity, Normal Forces etc. Consider a women on a swing: When is the tension on the rope largest ? And is it : (A) greater than (B) the same as (C) less than the force due to gravity acting on the woman Active Figure

23 Physics 207: Lecture 7, Pg 23 A match box car is going to do a loop-the-loop of radius r. What must be its minimum speed, v, at the top so that it can manage the loop successfully ? Loop-the-loop 1

24 Physics 207: Lecture 7, Pg 24 To navigate the top of the circle its tangential velocity, v, must be such that its centripetal acceleration at least equals the force due to gravity. At this point N, the normal force, goes to zero. Loop-the-loop 1 F c = - ma = - mg = - mv 2 /r v = (gr) 1/2 mgmg v

25 Physics 207: Lecture 7, Pg 25 The match box car is going to do a loop the loop. If the speed at the bottom is v B, what is the normal force, N, at that point? Hint: The car is constrained to the track. Loop-the-loop 3 mgmg v N

26 Physics 207: Lecture 7, Pg 26 Lecture 7, Example 7 Accelerated Reference Frames You are a passenger in a car and not wearing your seatbelt. Without increasing or decreasing speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is a correct description of the situation ? (A) Before and after the collision there is a rightward force pushing you into the door. (B) Starting at the time of the collision, the door exerts a leftward force on you. (C) Both of the above. (D) Neither of the above.

27 Physics 207: Lecture 7, Pg 27 Air Resistance and Drag l So far we’ve “neglected air resistance” in physics  Can be difficult to deal with l Affects projectile motion  Friction force opposes velocity through medium  Imposes horizontal force, additional vertical forces  Terminal velocity for falling objects l Dominant energy drain on cars, bicyclists, planes l This issue has been with us a very long time….

28 Physics 207: Lecture 7, Pg 28 Recapping Agenda: Assignment: l WebAssign Problem Set 3 due Tuesday midnight l MidTerm Thursday, Oct. 5, Chapters 1-6, 90 minutes, 7-8:45 PM l NOTE: Assigned Rooms are 105 and 113 Psychology l Chapter 5 (Forces and Newton’s Laws)   Static Friction  Problem exercise Chapter 6 (Circular Motion and Other Applications)  Friction (a external force that opposes motion)  Uniform and non-uniform circular motion  Accelerated Frames  Resistive Forces


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